A spherical ball of density - and radius 0-003 m is dropped into a tube containing a viscous fluid up to the 0 cm mark as shown in the figure- Viscosity of the fluid -1-26 N-s-m2 and its density -L -2- 1260 kg- m3 Assume the ball reaches a terminal speed at 10 cm mark- The time taken by the ball to travel the distance between the 10 cm and 20 cm mark is g - 10m-s 2

The correct option is D 5 s v_T=2/9r^2 ( ρ σ nbsp; )g/η nbsp; nbsp; =2× ( 0.003 )^2 ( 1260× 2 1260 )× 10/9× 1.26 nbsp; nbsp; ...

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Byju's Answer

Standard XII

Physics

Reynold's Number

A spherical b...

Question

A spherical ball of density $ρ$ and radius $0.003 m$ is dropped into a tube containing a viscous fluid up to the $0 c m$ mark as shown in the figure. Viscosity of the fluid $= 1.26 N - s / m^{2}$ and its density $ρ_{L} = \frac{ρ}{2} = 1260 k g / m^{3}$ Assume the ball reaches a terminal speed at $10 c m$ mark. The time taken by the ball to travel the distance between the $10 c m$ and $20 c m$ mark is $(g = 10 m / s^{2})$

2 s

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1 s

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0.5 s

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5 s

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Solution

The correct option is D $5 s$
$v_{T} = \frac{2}{9} \frac{r^{2} (ρ - σ) g}{η}$
$= \frac{2 \times {(0.003)}^{2} (1260 \times 2 - 1260) \times 10}{9 \times 1.26}$
$= 0.02$ m/s
$t = \frac{d}{v_{T}} = \frac{10 \times 10^{- 2}}{0.02} = 5$ s

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The correct option is D 5svT=29r2(ρ−σ)gη =2×(0.003)2(1260×2−1260)×109×1.26 =0.02 m/st=dvT=10×10−20.02=5 s

The correct option is D $5 s$
$v_{T} = \frac{2}{9} \frac{r^{2} (ρ - σ) g}{η}$
$= \frac{2 \times {(0.003)}^{2} (1260 \times 2 - 1260) \times 10}{9 \times 1.26}$
$= 0.02$ m/s
$t = \frac{d}{v_{T}} = \frac{10 \times 10^{- 2}}{0.02} = 5$ s